Question

Integrate the function: $\sqrt {\sin2x}\cos 2x$

Answer 1

Let sin 2x = t

∴ 2cos 2x dx = dt

$\Rightarrow \int\sqrt {\sin2x} \;\cos 2xdx=\frac{1}{2}\int\sqrt t\;dt$

= $\frac{1}{2}\bigg(\frac{t^{\frac{3}{2}}}{\frac{3}{2}}\bigg)+C$

=$\frac{1}{3}t^{\frac{3}{2}}+C$

=$\frac{1}{3}(\sin2x)^{\frac{3}{2}}+C$